Problem: You have found the following ages (in years) of all 6 zebras at your local zoo: $ 22,\enspace 21,\enspace 13,\enspace 19,\enspace 16,\enspace 18$ What is the average age of the zebras at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Because we have data for all 6 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{22 + 21 + 13 + 19 + 16 + 18}{{6}} = {18.2\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $3.8$ years $14.44$ years $^2$ $21$ years $2.8$ years $7.84$ years $^2$ $13$ years $-5.2$ years $27.04$ years $^2$ $19$ years $0.8$ years $0.64$ years $^2$ $16$ years $-2.2$ years $4.84$ years $^2$ $18$ years $-0.2$ years $0.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{14.44} + {7.84} + {27.04} + {0.64} + {4.84} + {0.04}} {{6}} $ $ {\sigma^2} = \dfrac{{54.84}}{{6}} = {9.14\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{9.14\text{ years}^2}} = {3\text{ years}} $ The average zebra at the zoo is 18.2 years old. There is a standard deviation of 3 years.